Question

Consider the series.

Answer 1

Put 1

$\\ \rm\Rrightarrow \left(\dfrac{3^{1+1}}{3(1)+1}\right)$

$\\ \rm\Rrightarrow 3^2/4=9/4=2.2$

Put 2

$\\ \rm\Rrightarrow \left(\dfrac{3^{2+1}}{3(2)+1}\right)=3^3/7=27/7=3.8$

R is

• 3.8/2.2=19/11

Its greater than 1 .

• So diverge series.

Answer 2

Answer:

$r=\dfrac{12}{7}$

series diverges

Step-by-step explanation:

To find the common ratio (r) of a geometric series, divide the (n+1)th term by the nth term.

When n = 1:

$a_1=\dfrac{3^{1+1}}{3(1)+1}=\dfrac{3^2}{4}=\dfrac94$

When n =2:

$a_2=\dfrac{3^{2+1}}{3(2)+1}=\dfrac{3^3}{7}=\dfrac{27}{7}$

Therefore,

$r=\dfrac{a_2}{a_1}=\dfrac{\frac{27}{7}}{\frac{9}{4}}=\dfrac{12}{7}$

A series that converges has a finite limit.  If |r| < 1, then the series will converge.

A series that diverges means either the partial sums have no limit or approach infinity.  If |r| > 1 then the series diverges.

Therefore, as the limit of the series approaches infinity and it's r value is greater than 1, the series diverges.